Question
Answer
$$\dfrac{4x-8}{x^2-4x+4}=\dfrac{4}{x-2}$$
Explanation
| $$ \begin{aligned}\frac{4x-8}{x^2-4x+4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4}{x-2}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{4x-8}{x^2-4x+4} $ to $ \dfrac{4}{x-2} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-2}$. $$ \begin{aligned} \frac{4x-8}{x^2-4x+4} & =\frac{ 4 \cdot \color{blue}{ \left( x-2 \right) }}{ \left( x-2 \right) \cdot \color{blue}{ \left( x-2 \right) }} = \\[1ex] &= \frac{4}{x-2} \end{aligned} $$ |
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